What is the integral of x(e^x-1) ?

The integral of x(e^x-1) is e^xx-x^2-e^x+(x^2)/2+C

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integral of x(e^x-1)

\int x (e^{x} - 1) d x

e^{x} x - x^{2} - e^{x} + \frac{x ^{2}}{2} + C

Solution steps

\int x (e^{x} - 1) d x

Apply Integration By Parts

: x (e^{x} - x) - \int e^{x} - x d x

= x (e^{x} - x) - \int e^{x} - x d x

\int e^{x} - x d x = e^{x} - \frac{x ^{2}}{2}

= x (e^{x} - x) - (e^{x} - \frac{x ^{2}}{2})

Simplify

x (e^{x} - x) - (e^{x} - \frac{x ^{2}}{2}) : e^{x} x - x^{2} - e^{x} + \frac{x ^{2}}{2}

= e^{x} x - x^{2} - e^{x} + \frac{x ^{2}}{2}

Add a constant to the solution

= e^{x} x - x^{2} - e^{x} + \frac{x ^{2}}{2} + C

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What is the integral of x(e^x-1) ?
The integral of x(e^x-1) is e^xx-x^2-e^x+(x^2)/2+C